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\title{\heiti\zihao{2} 习题14.3}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{确定下列函数的定义域}
\subsection{$f(x, y)=\sqrt{1-x^{2}-y^{2}}$}
\textbf{解}\quad
定义域为$\left\{ (x,y)\mid x^2+y^2\leqslant 1 \right\}$

\subsection{$f(x, y)=\sqrt{x-y}$}
\textbf{解}\quad
定义域为$\left\{(x,y)\mid x\geqslant y\right\}$

\subsection{$f(x, y, z)=\arccos \dfrac{x^{2}+y^{2}}{z}$}
\textbf{解}\quad
定义域为$\left\{(x,y)\mid -1\leqslant\dfrac{x^2+y^2}{z}\leqslant 1\right\}$

\subsection{$f(x, y, z)=\ln (x+y+z)$}
\textbf{解}\quad
定义域为$\left\{(x,y,z)\mid x+y+z>0\right\}$

\subsection{$f(x, y)=\arcsin \dfrac{y}{x}$}
\textbf{解}\quad
定义域为$\left\{(x,y)\mid -1\leqslant \dfrac{y}{x}\leqslant 1\right\}$

\subsection{$f(x, y)=\sqrt{1-x^{2}}+\sqrt{y^{2}-1}$}
\textbf{解}\quad
定义域为$\left\{(x,y)\mid x^2\leqslant 1 \text{且}y^2\geqslant 1\right\}$

\section{计算下列极限}
\subsection{$\lim\limits_{(x,y)\rightarrow(0,0)}\dfrac{\sin xy}{x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim\limits_{(x,y)\rightarrow(0,0)}\dfrac{\sin xy}{x} & =\lim\limits_{(x,y)\rightarrow(0,0)}\dfrac{xy}{x}\cdot\dfrac{\sin xy}{xy} = 0
	\end{aligned}
$$

\subsection{$\lim\limits_{(x,y)\rightarrow(+\infty,+\infty)}\left(\dfrac{xy}{x^2+y^2}\right)^{x^2}$}
\textbf{解}\quad
$$
	\begin{aligned}
		0 & \leqslant\lim\limits_{(x,y)\rightarrow(+\infty,+\infty)}\left(\dfrac{xy}{x^2+y^2}\right)^{x^2}             \\
		  & \leqslant\lim\limits_{(x,y)\rightarrow(+\infty,+\infty)}\left(\dfrac{1}{2}\right)^{x^2}(\text{均值不等式}) \\
		  & \leqslant 0
	\end{aligned}
$$
所以由夹逼准则知极限为$0$.

\subsection{$\lim\limits_{(x,y)\rightarrow(+\infty,0)}\left(1+\dfrac{2}{x}\right)^{x^2/(x+y)}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim\limits_{(x,y)\rightarrow(+\infty,0)}\left(1+\dfrac{2}{x}\right)^{x^2/(x+y)} & =\lim\limits_{(x,y)\rightarrow(+\infty,0)}\left(1+\dfrac{2}{x}\right)^{\dfrac{x}{2}\cdot\dfrac{2x}{(x+y)}} \\
		                                                                                 & =e^2
	\end{aligned}
$$

\subsection{$\lim\limits_{(x, y) \rightarrow(0,0)} \dfrac{x^{2}+y^{2}}{|x|+|y|}$}
\textbf{解}\quad
由Cauchy不等式:
$$
	\begin{aligned}
		\lim\limits_{(x, y) \rightarrow(0,0)} \dfrac{(|x|+|y|)^2}{2(|x|+|y|)}\leqslant\lim\limits_{(x, y) \rightarrow(0,0)} \dfrac{(|x|+|y|)^2}{|x|+|y|}\leqslant\lim\limits_{(x, y) \rightarrow(0,0)} \dfrac{(|x|+|y|)^2}{(|x|+|y|)}
	\end{aligned}
$$
所以由夹逼准则知极限为$0$.

\subsection{$\lim\limits_{(x, y) \rightarrow(+\infty,+\infty)}\left(x^{2}+y^{2}\right) \mathrm{e}^{-(x+y)}$}
\textbf{解}\quad
由于$x,y$趋于正无穷,从而只需考虑二者都是正数的情况.
$$
	\begin{aligned}
		0\leqslant \lim\limits_{(x, y) \rightarrow(+\infty,+\infty)}\left(x^{2}+y^{2}\right) \mathrm{e}^{-(x+y)}\leqslant\lim\limits_{(x, y) \rightarrow(+\infty,+\infty)}\dfrac{(x+y)^2}{\mathrm{e}^{(x+y)}} = 0
	\end{aligned}
$$
由夹逼准则知极限为$0$.

\subsection{$\lim\limits_{(x, y) \rightarrow(0,1)} \dfrac{1-x y}{x^{2}+y^{2}}$}
\textbf{解}\quad
$$
	\lim\limits_{(x, y) \rightarrow(0,1)} \dfrac{1-x y}{x^{2}+y^{2}} = \dfrac{1}{1} = 1
$$

\section{讨论下列函数在 $(0,0)$ 点的重极限与累次极限}
\subsection{$f(x, y)=y \sin \dfrac{1}{x}$}
\textbf{解}\quad
$y\sin \dfrac{1}{x}\leqslant y$,从而重极限存在.\par
由于$\lim\limits_{x\rightarrow 0}\sin \dfrac{1}{x}$不存在,从而$\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}y\dfrac{1}{x}$不存在.\par
但$\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}y\sin \dfrac{1}{x} = 0$,所以这个累次极限存在.

\subsection{$f(x, y)=\left(x^{2}+y^{2}\right)^{x^{2} y^{2}}$}
\textbf{解}\quad
$$
	\lim\limits_{(x, y) \rightarrow(0,0)}(x^2+y^2)\ln (x^2+y^2) = 0
$$
从而
$$
	\lim\limits_{(x, y) \rightarrow(0,0)}(x^2+y^2)^{x^2y^2} = \mathrm{e}^0=1
$$
累次极限:$\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0} \left(x^{2}+y^{2}\right)^{x^{2} y^{2}}= \lim\limits_{y\rightarrow 0} (y^2)^0= 1$.由$x,y$对称性可知累次极限都存在.

\subsection{$f(x, y)=\dfrac{\sin (x y)}{y}$}
\textbf{解}\quad
重极限:
$$
	\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{\sin (x y)}{y} = \lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{x y}{y} = 0
$$
累次极限:
$$
	\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}\dfrac{\sin (x y)}{y} = \lim\limits_{y\rightarrow 0}\dfrac{0}{y}=0
$$
以及
$$
	\begin{aligned}
		\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{\sin (x y)}{y} & = \lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{xy}{y}\cdot\dfrac{\sin(xy)}{xy} \\
		                                                                              & =\lim\limits_{x\rightarrow 0}x=0
	\end{aligned}
$$
从而两个累次极限都存在.

\subsection{$f(x, y)=\dfrac{\sin x y}{\sqrt{x^{2}+y^{2}}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{\sin x y}{\sqrt{x^{2}+y^{2}}} & =\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{xy}{\sqrt{x^2+y^2}}                         \\
		                                                                          & = \lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{1}{\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}}}=0
	\end{aligned}
$$
累次极限:
$$
	\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{\sin x y}{\sqrt{x^{2}+y^{2}}}=\lim\limits_{x\rightarrow 0}\dfrac{0}{x^2}=0
$$
由$x,y$的对称性可知累次极限存在.

\subsection{$f(x, y)=\dfrac{x^{2}+y^{2}}{\sqrt{1+x^{2}+y^{2}}-1}$}
\textbf{解}\quad
重极限:
$$
	\begin{aligned}
		\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{x^{2}+y^{2}}{\sqrt{1+x^{2}+y^{2}}-1} & =\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{x^2+y^2}{\dfrac{x^2+y^2}{2}}=2
	\end{aligned}
$$
累次极限:
$$
	\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0} \dfrac{x^{2}+y^{2}}{\sqrt{1+x^{2}+y^{2}}-1}= \lim\limits_{x\rightarrow 0}\dfrac{x^2}{\sqrt{1+x^2}-1}=\lim\limits_{x\rightarrow 0}\dfrac{x^2}{\dfrac{x^2}{2}}=2
$$
由$x,y$的对称性可知累次极限存在.
\subsection{$f(x, y)=\dfrac{\ln \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$}
\textbf{解}\quad
重极限:
$$
	\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{\ln \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} = \lim_{u\rightarrow 0}\dfrac{\ln u}{u} = 0
$$
累次极限:
$$
	\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{\ln \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} = \lim\limits_{x\rightarrow 0}\dfrac{\ln x^2}{x^2} = 0
$$
由$x,y$的对称性可知累次极限存在.

\subsection{$f(x, y)=\dfrac{x^{2}+y^{2}}{x^{2} y^{2}+(x-y)^{2}}$}
\textbf{解}\quad
令$y=kx$,则
$$
	\lim\limits_{x\rightarrow 0}\dfrac{x^2+k^2x^2}{k^2x^4+(1-k)^2x^2}=\dfrac{k^2+1}{(1-k)^2}
$$
与$k$有关,所以重极限不存在.
$$
	\begin{aligned}
		\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{\ln \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} = \lim\limits_{x\rightarrow 0}1 = 1 \\
		\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}\dfrac{\ln \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} = \lim\limits_{y\rightarrow 0}1 = 1
	\end{aligned}
$$
所以累次极限存在且相等.

\subsection{$f(x, y)=\dfrac{x+y}{x-y}$}
\textbf{解}\quad
两个累次极限:
$$
	\begin{aligned}
		\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{x+y}{x-y} & =\lim\limits_{x\rightarrow 0}\dfrac{x}{x} = 1 \\
		\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}\dfrac{x+y}{x-y} & = \dfrac{y}{-y} = -1
	\end{aligned}
$$
累次极限不相等,重极限不存在.

\subsection{$f(x, y)=|y|^{|x|}$}
\textbf{解}\quad
不妨$x,y>0$.令$y=\dfrac{1}{k^{1/x}},k>1$,从而重极限:
$$
	\lim\limits_{(x, y) \rightarrow(0,0)}|y|^{|x|} = \dfrac{1}{k}
$$
与$k$有关,从而重极限不存在.
$$
	\begin{aligned}
		\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}|y|^{|x|}=\lim\limits_{x\rightarrow 0}0^{|x|}=0 \\
		\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}|y|^{|x|} = \lim\limits_{y\rightarrow 0}1 = 1
	\end{aligned}
$$
从而累次极限都存在.
\subsection{$f(x, y)=\dfrac{\ln (1+x y)}{x+\tan y}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{\ln (1+x y)}{x+\tan y} & =\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{xy}{x+\tan y}                                           \\
		                                                                   & =\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{x}{\dfrac{x}{y}+1}                                      \\
		                                                                   & =\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{xy}{x+y}                                                \\
		                                                                   & =\lim\limits_{r \rightarrow 0}r\dfrac{\sin \theta\cos \theta}{\sqrt{2}\sin(\theta + \dfrac{\pi}{4})}
	\end{aligned}
$$
取逼近曲线$\theta = -\dfrac{\pi}{4}+kr$,从而极限为$\dfrac{1}{2k}$,与$k$有关,从而重极限不存在.\par
累次极限:
$$
	\begin{aligned}
		\lim\limits_{x\rightarrow 0}\lim\limits_{y\rightarrow 0}\dfrac{\ln (1+x y)}{x+\tan y} & =\lim\limits_{x\rightarrow 0}0 = 0 \\
		\lim\limits_{y\rightarrow 0}\lim\limits_{x\rightarrow 0}\dfrac{\ln (1+x y)}{x+\tan y} & =\lim\limits_{y\rightarrow 0}0 = 0
	\end{aligned}
$$
从而累次极限都存在且相等.\par
\textcolor{blue}{\textbf{注:}重极限存在,则累次极限不存在的唯一可能情况是最后一层极限之前的极限过程不存在,而不可能是前面的过程极限存在而最后一个极限过程不存在.}

\section{讨论下列二元函数在其定义域内的连续性}
\subsection{$f(x, y)=\left\{\begin{array}{cc}\dfrac{x^{2} y^{2}}{x^{2}+y^{2}}, & x^{2}+y^{2} \neq 0, \\ 0, & x^{2}+y^{2}=0\end{array}\right.$}
\textbf{解}\quad
重极限:
$$
	\begin{aligned}
		\lim\limits_{(x, y) \rightarrow(0,0)}\dfrac{x^{2} y^{2}}{x^{2}+y^{2}} & =\lim\limits_{r \rightarrow 0}r^2\sin^2 \theta\cos^2\theta = 0
	\end{aligned}
$$
显然由与连续函数四则运算后得到的都是连续函数,又在$0$处连续,从而函数在定义域内连续.

\subsection{$f(x, y)=\left\{\begin{array}{cl}\dfrac{\ln (1+x y)}{x}, & x \neq 0 \\ y, & x=0\end{array}\right.$}
\textbf{解}\quad
$f(x, y)$ 的定义域是 $\{(x, y) \mid x y>-1\}$ 且 $f(x, y)$ 在 $x \neq 0$ 处是连续的, 所 以只须证明 $f(x, y)$ 作为二元函数在 $y$ 轴上的每一点处连续. 以下分两种情况讨论.

(1) 在点 $(0,0)$. 由于 $f(0,0)=0$, 而当 $x \neq 0$ 时
$$
	f(x, y)=\dfrac{\ln (1+x y)}{x}=\left\{\begin{array}{ll}
		0,                              & y=0,       \\
		y \ln (1+x y)^{\dfrac{1}{x y}}, & y \neq 0 .
	\end{array}\right.
$$
又由于
$$
	\lim _{(x, y) \rightarrow(0,0)} \ln (1+x y)^{\dfrac{1}{x y}}=1
$$
从而 $\exists \delta_{1}>0$, 当 $0<|x|<\delta_{1}, 0<|y|<\delta_{1}$ 时
$$
	\left|\dfrac{\ln (1+x y)}{x}\right| \leqslant|y| \cdot\left|\ln (1+x y)^{\dfrac{1}{x y}}\right| \leqslant 2|y|
$$
由 $f(x, y)$ 的表达式知只要 $|x|<\delta_{1},|y|<\delta_{1}$, 无论 $x=0$ 还是 $x \neq 0$ 都有
$$
	|f(x, y)| \leqslant 2|y| .
$$
所以
$$
	\lim _{(x, y) \rightarrow(0,0)} f(x, y)=0=f(0,0)
$$

(2) 在点 $\left(0, y_{0}\right), y_{0} \neq 0$.当$x \neq 0$时
$$
	\begin{aligned}
		\left|f(x, y)-f\left(0, y_{0}\right)\right| & =\left|y \ln (1+x y)^{\dfrac{1}{x y}}-y_{0}\right|                                   \\
		                                            & =\left|y\left[\ln (1+x y)^{\dfrac{1}{x y}}-1\right]+\left(y-y_{0}\right)\right|      \\
		                                            & \leqslant|y| \cdot\left|\ln (1+x y)^{\dfrac{1}{x y}}-1\right|+\left|y-y_{0}\right| .
	\end{aligned}
$$
而当 $x=0$ 时
$$
	\left|f(x, y)-f\left(0, y_{0}\right)\right|=\left|y-y_{0}\right|
$$
注意到当 $y_{0} \neq 0$ 时
$$
	\lim _{(x, y) \rightarrow\left(0, y_{0}\right)} \ln (1+x y)^{\dfrac{1}{x y}}=1
$$
结合上述各式得
$$
	\lim _{(x, y) \rightarrow\left(0, y_{0}\right)}\left(f(x, y)-f\left(0, y_{0}\right)\right)=0
$$
所以 $f(x, y)$ 在点 $\left(0, y_{0}\right)$ 处连续, 从而在其定义域上是连续的.


\subsection{$f(x, y)=\left\{\begin{array}{cl}\dfrac{1}{\sqrt{x^{2}+y^{2}}}, & x^{2}+y^{2} \neq 0, \\ 0, & x^{2}+y^{2}=0 .\end{array}\right.$}
\textbf{解}\quad
$(x,y)\rightarrow(0,0)$的时候,$\dfrac{1}{\sqrt{x^2+y^2}}=+\infty$,从而极限不存在,所以在$0$处不连续.

\section{举例说明存在 $[a, b] \times[c, d]$ 上的函数 $f(x, y)$, 对于任意给定的 $x_{0} \in[a, b]$, 函数 $f\left(x_{0}, y\right)$ 在 $[c, d]$ 上连续, 对于任意给定的 $y_{0} \in[c, d]$, 函数 $f\left(x, y_{0}\right)$ 在 $[a, b]$ 上连续,但函 $f(x, y)$ 在 $[a, b] \times[c, d]$ 上不连续.}
$$
	f(x, y) = \dfrac{xy}{x+y}
$$
定义$f(0,0)=0$.由上文的讨论,对$(x,y)\in[0,1]\times [0,1]$,累次极限都存在且相等,显然函数在$x,y$的定义域内关于$x,y$分别都是连续的,但函数本身不存在$(0,0)$处的重极限,所以函数在$(0,0)$处不连续.

\section{设二元函数 $f(x, y)$ 在集合 $D \in \mathbb{R}^{2}$ 上对于每一个自变量 $x$ 和 $y$ 都连续. 证明: 当下列 件之一满足时, $f(x, y)$ 是 $D$ 上的连续函数.\\
(1) 对任意给定的 $x$, 函数 $f(x, y)$ 关于 $y$ 单调;\\
(2) 关于变量 $y$ 满足 Lipschitz 条件,即存在常数 $L$, 对任意 \\$\left(x, y^{\prime}\right),\left(x, y^{\prime \prime}\right) \in D$,都有$$
	\left|f\left(x, y^{\prime}\right)-f\left(x, y^{\prime \prime}\right)\right| \leqslant L\left|y^{\prime}-y^{\prime \prime}\right|
$$
(3) $f(x, y)$ 在 $D$ 上关于 $x$ 连续,且关于 $y$ 一致连续,即对 $\forall x_{0}$ 和 $\forall \varepsilon>0$, 都有 $\exists \delta>0$, 当 $\left|x-x_{0}\right|<\delta$ 时,对任意 $y,(x, y),\left(x_{0}, y\right) \in D$, 都有
$$
	|f(x, y)-f\left(x_{0}, y\right) |<\varepsilon
$$}
\subsection{}
\begin{proof}
	取矩形邻域$[x_0-\delta_1,x_0+\delta_1]\times[y_0-\delta_2,y_0+\delta_2]$.\par
	考虑$f(x_0,y)$,其关于$y$连续,从而$\forall\varepsilon>0,\exists\delta_2>0$,s.t.$|f(x_0,y_0\pm\delta_2)-f(x_0,y_0)|<\varepsilon$.\par
	考虑$f(x,y_0)$,其关于$x$连续,从而$\forall\varepsilon>0,\exists\delta_1>0$,s.t.$|f(x_0\pm\delta_1,y_0)-f(x_0,y_0)|<\varepsilon$.\par
	再由$f$关于$y$的单调性可知对于$\forall(x,y)$在$(x_0,y_0)$的方形邻域内,都有$|f(x,y)-f(x_0,y_0)|\leqslant|f(x,y_0\pm\delta_2)-f(x_0,y_0)|$.所以再由上面的$f$关于$x,y$连续的性质可知:
	$$
		|f(x,y_0\pm\delta_2)-f(x_0,y_0)|\leqslant|f(x,y_0\pm\delta_2)-f(x,y_0)|+|f(x,y_0)-f(x_0,y_0)|\leqslant\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon
	$$
\end{proof}

\subsection{}
\begin{proof}
	$$
		|f(x,y)-f(x_0,y_0)|\leqslant|f(x,y)-f(x_0,y)|+|f(x_0,y)-f(x_0,y_0)|
	$$
	由于$f$关于$x$连续,从而上式RHS第一项可取$\delta$,使其小于$\forall
		\dfrac{\varepsilon}{2}>0$,第二项由于满足Lipschitz条件,故可取$\delta'$,使其小于$\forall\dfrac{\varepsilon}{2}>0$.所以可得结论.
\end{proof}

\subsection{}
\begin{proof}
	$\forall\left(x_{0}, y_{0}\right), \forall \varepsilon>0, \exists \delta_{1}=\delta_{1}\left(\varepsilon, x_{0}\right)>0$ (与 $y$ 无关 )\par
	当 $\left|x-x_{0}\right|<\delta_{1}$ 时,对一切 $y$ 有 $\left|f(x, y)-f\left(x_{0}, y\right)\right|<\dfrac{\varepsilon}{2} .$ 又因 $\left(x_{0}, y_{0}\right)$ 处 $f\left(x_{0}, y\right)$ 对 $y$ 连续,故对此 $\varepsilon>$
	$0, \exists \delta_{2}>0$ 使得 $\left|y-y_{0}\right|<\delta_{2}$ 时, 有 $\left|f\left(x_{0}, y\right)-f\left(x_{0}, y_{0}\right)\right|<\dfrac{\varepsilon}{2}$.\par
	取 $\delta=\min \left\{\delta_{1}, \delta_{2}\right\}$,则 $\left|x-x_{0}\right|<\delta,\left|y-y_{0}\right|<\delta$ 时, 有
	$$
		\begin{aligned}
			\left|f(x, y)-f\left(x_{0}, y_{0}\right)\right| & \leqslant\left|f(x, y)-f\left(x_{0}, y\right)\right|+\left|f\left(x_{0}, y\right)-f\left(x_{0}, y_{0}\right)\right| \\
			                                                & <\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon
		\end{aligned}
	$$
\end{proof}
\end{document}
\subsection{}
\textbf{解}\quad